Processing Math: Done
Solution 2.3:2b
From Förberedande kurs i matematik 1
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| - | {{ | + | The first step when we solve the second-degree equation is to complete the square on the left-hand side |
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| - | {{ | + | {{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}} |
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| + | The equation can now be written as | ||
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| + | {{Displayed math||<math>(y+1)^{2} = 16</math>}} | ||
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| + | and has, after taking the square root, the solutions: | ||
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| + | :*<math>y+1 = \sqrt{16} = 4\,\textrm{,}\ </math> which gives <math>y=-1+4=3\,\textrm{,}</math> | ||
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| + | :*<math>y+1 = -\sqrt{16} = -4\,\textrm{,}\ </math> which gives <math>y=-1-4=-5\,\textrm{.}</math> | ||
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| + | A quick check shows that <math>y=-5</math> and <math>y=3</math> satisfy the equation: | ||
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| + | :*''y'' = -5: <math>\ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}</math> | ||
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| + | :*''y'' = 3: <math>\ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}</math> | ||
Current revision
The first step when we solve the second-degree equation is to complete the square on the left-hand side
The equation can now be written as
and has, after taking the square root, the solutions:
y+1= which gives
16=4, y=−1+4=3,
y+1=− which gives16=−4, y=−1−4=−5.
A quick check shows that
- y = -5:
LHS=(−5)2+2 
(−5)−15=25−10−15=0=RHS,
- y = -5:
- y = 3:
LHS=32+2 3−15=9+6−15=0=RHS.
- y = 3:
